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AnnZ [28]
2 years ago
10

A balloon containing 0. 500 mol Ar at 0. 00°C and 65. 0 kPa pressure is expanded by adding more argon. Formula: PV = nRT (R = 8.

317 L•kPa/mol•K) What is the original volume of the gas? L.
Mathematics
1 answer:
Nuetrik [128]2 years ago
8 0

Using an ideal gas equation to solve the problem. Then the original volume is 17.67 liters.

<h3>What is an ideal gas equation?</h3>

An ideal gas is a theoretical composed of a set of randomly moving gas particles that interact only through elastic collision.

Given

A balloon containing 0.500 mol at 0. 00°C and 65.0 kPa pressure are expanded by adding more argon.

Pressure (P) = 65 kPa

Mole (n) = 0.5 mol

Temperature (T) = 0 °C = 273.15 K

Universal gas constant (R) = 8.317 L.kPa/mol.K

We know the ideal gas equation

\rm PV =nRT\\\\V \ \ = \dfrac{nRT}{P}

Put the values, we have

\rm V = \dfrac{0.5*8.317*273.15}{65}\\\\V = 17.67

Thus, the original volume is 17.67 liters.

More about the ideal gas equation link is given below.

brainly.com/question/4147359

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Answer:

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Step-by-step explanation:

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Hence, Ms gill pay for all three banners=$896

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Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that th
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Answer:

Step-by-step explanation:

(a)

The bid should be greater than $10,000 to get accepted by the seller. Let bid x be a continuous random variable that is uniformly distributed between

$10,000 and $15,000

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The interval of the provided bidding is [$10,000,$12,000]. The probability is calculated as,

\begin{array}{c}\\P\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{12000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{12000}^{15000}\\\end{array}

=1- \frac{[15000-12000]}{5000}\\\\=1-0.6\\\\=0.4

(b)  The interval of the accepted bidding is [$10,000,$15,000], where b = $15,000 and a =$10,000. The interval of the given bidding is [$10,000,$14,000].

\begin{array}{c}\\P\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{15000 - 10000}}} dx\\\\ = 1 - \int\limits_{14000}^{15000} {\frac{1}{{5000}}} dx\\\\ = 1 - \frac{1}{{5000}}\left[ x \right]_{14000}^{15000}\\\end{array} P(X14000)

=1- \frac{[15000-14000]}{5000}\\\\=1-0.2\\\\=0.8

(c)

The amount that the customer bid to maximize the probability that the customer is getting the property is calculated as,  

The interval of the accepted bidding is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The interval of the given bidding is [$10,000,$15,000].

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(d)  The amount that the customer bid to maximize the probability that the customer is getting the property is $15,000, set by the seller. Another customer is willing to buy the property at $16,000.The bidding less than $16,000 getting considered as the minimum amount to get the property is $10,000.

The bidding amount less than $16,000 considered by the customers as the minimum amount to get the property is $10,000, and greater than $16,000 will depend on how useful the property is for the customer.

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